## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 46

#### Answer

$\displaystyle \frac{1}{2p^{2}}$

#### Work Step by Step

$\displaystyle \frac{8p^{-3}(4p^{2})^{-2}}{p^{-5}} \qquad$ ....... use $(ab)^{m}=a^{m} b^{m}$ $=\displaystyle \frac{8p^{-3}\cdot 4^{-2}(p^{2})^{-2}}{p^{-5}}$ ....... use $(a^{m})^{n}=a^{mn}$, group like terms $=8\displaystyle \cdot 4^{-2}\cdot\frac{p^{-3}\cdot p^{-4}}{p^{-5}}$ ....... use $a^{m}\cdot a^{n}=a^{m+n} ,\quad \displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$ .......recognize: $8=2^{3},\quad 4=2^{2}$ $=(2^{3})(2^{2})^{-2}\cdot p^{-3-4-(-5)}\qquad$ ....... use $(a^{m})^{n}=a^{mn}$ $=(2^{3})(2^{-4})\cdot p^{-2}\qquad$ ....... use $a^{m}\cdot a^{n}=a^{m+n}$ $=2^{3-4}p^{-2}$ $=2^{-1}p^{-2}\qquad$ ....... use $a^{-n}=\displaystyle \frac{1}{a^{n}}=(\frac{1}{a})^{n}$ $= \displaystyle \frac{1}{2p^{2}}$

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